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NEET 2027 · Physics · Class 11 · Ch. 1 Units and Measurement

Least Count — Vernier & Screw Gauge

Vernier constant, pitch, zero error and the reading formula. Another syllabus-gap topic: barely present in the rationalised NCERT text, repeatedly present in NEET papers.

PRIORITY 2TOPIC 5 OF 1520 QUESTIONS4 NCERT EXERCISESSYLLABUS GAP TOPIC

01Syllabus warning — read this first

Instruments appear in the exercises, not in the chapter body

The rationalised NCERT chapter contains no section on measuring instruments. Vernier callipers and screw gauges survive only inside Exercises 1.6, 1.7 and 1.8 — four questions with no supporting text.

The NTA syllabus for Unit I nevertheless lists least count explicitly, and NEET has asked instrument questions repeatedly, including a screw-gauge pitch item and a vernier constant item in the 2024 re-examination and a screw-gauge item in 2022.

Conclusion: this topic must be learned from outside the textbook. Everything on this page is examinable.

02Concept in plain language

Least count is the smallest change an instrument can actually show you. It is the size of one step on the finest scale — and therefore the built-in uncertainty of any single reading.

A metre scale is marked every millimetre, so its least count is 1 mm. Anything thinner than that is guesswork. Both the vernier callipers and the screw gauge exist to beat that limit, and they do it by two quite different tricks.

The vernier trick: a deliberate mismatch

The sliding vernier scale is made slightly smaller than the main scale. If ten vernier divisions span nine millimetres, each vernier division is 0.9 mm — a tenth of a millimetre short of a main division. Slide the vernier along and exactly one of its lines will fall into step with a main scale line. The number of that line tells you how many tenths of a millimetre past the whole number you are.

That deliberate mismatch is the least count: LC = 1 MSD − 1 VSD, which simplifies to MSD / N.

The screw gauge trick: turn rotation into translation

A screw with a fine thread converts a large rotation into a tiny advance. One full turn moves the spindle forward by the pitch. Wrap a hundred divisions around that rotation and each division corresponds to one hundredth of the pitch: LC = pitch / N.

Both instruments read the same way

reading = coarse scale reading + (fine division number × least count)

Then subtract the zero error. That single line, plus the two least-count formulae, is the whole topic.

03The two instruments

VERNIER SCALE · 10 VERNIER DIVISIONS SPAN 9 MAIN SCALE DIVISIONS0510main0123456789106th division coincidesLC = 1 MSD − 1 VSD = 1 mm − 0.9 mm = 0.1 mmReading = main scale reading + (coinciding division × LC) = 3 mm + 6 × 0.1 mm = 3.6 mmOnly one vernier line ever lines up exactly — that line number is the fractional part.
Fig. 1 — A ten-division vernier sliding to a reading of 3.6 mm. Exactly one vernier line ever coincides, and its number is the fractional part of the reading.
SCREW GAUGE · ONE FULL ROTATION ADVANCES THE SPINDLE BY ONE PITCH anvil spindle 0 1 2 3 pitch scale (mm) reference line circular scale (100 divisions) Pitch = distance advanced in one full rotation. LC = pitch ÷ number of circular divisions = 1 mm ÷ 100 = 0.01 mm. Reading = pitch-scale reading + (circular division at the reference line × LC).
Fig. 2 — One full rotation of the circular scale advances the spindle by one pitch. The least count is that pitch shared among the circular divisions.
ZERO ERROR · READ WITH THE JAWS FULLY CLOSED, BEFORE ANY MEASUREMENT 0 no zero error correction = 0 +5 positive zero error zero below the line → subtract +5 × 0.01 = +0.05 mm 96 negative zero error zero above the line → add −(100−96) × 0.01 = −0.04 mm True reading = observed reading − zero error. The subtraction of a negative error is what turns it into an addition.
Fig. 3 — Zero error, read with the jaws fully closed and before any measurement. A reading just above zero is a positive error; a reading near the top of the scale is a negative one.

04Formula sheet

Least count
Metre scale
LC = 1 mm
= 0.1 cm
Vernier callipers
LC = 1 MSD − 1 VSD = MSD / N
N vernier divisions spanning (N − 1) main divisions
Vernier, N = 10, MSD = 1 mm
LC = 0.1 mm = 0.01 cm
the standard school instrument
Vernier, N = 20, MSD = 1 mm
LC = 0.05 mm = 0.005 cm
NCERT Exercise 1.6(a)
Screw gauge pitch
pitch = (distance advanced) / (number of rotations)
extract this first whenever rotations are mentioned
Screw gauge
LC = pitch / N
N = number of circular scale divisions
Screw gauge, 1 mm pitch, 100 div
LC = 0.01 mm = 10 μm
the standard instrument
Spherometer
LC = pitch / N
same structure; used for surfaces and plates
Reading and correcting
Total reading
R = MSR + (n × LC)
n is the coinciding fine-scale division
Zero error, positive
ZE = + n × LC
zero of the fine scale sits past the reference line
Zero error, negative
ZE = −(N − n) × LC
closed-jaw reading near the top of the circular scale
Correction
true = observed − ZE
so a negative zero error is effectively added
Single-reading uncertainty
ΔR = LC
one full least count, unless told otherwise
Percentage error
(LC / R) × 100

05Zero error in full

Zero error is the reading an instrument shows when it ought to show nothing. It is a systematic error: identical in size and sign on every measurement, and completely immune to averaging. The only cure is to measure it and subtract it.

the sign of the zero error decides the operation — get it right before touching the arithmetic
CaseWhat you see with the jaws closedZero errorWhat to do
No zero errorFine-scale zero exactly on the reference line0nothing
PositiveFine-scale zero has moved past the reference line; a small division number such as 3 or 5 lines up+ n × LCsubtract it from every reading
NegativeFine-scale zero has not yet reached the reference line; a large division number such as 96 lines up−(N − n) × LCadd its magnitude to every reading

How to tell positive from negative in one glance

Look at the number that coincides when the jaws are closed. A small number (2, 3, 5) means the scale has gone slightly too far — positive error, subtract. A large number close to N (95, 96, 98) means the scale is slightly short of a full turn — negative error, add.

The arithmetic then follows: true = observed − zero error, and subtracting a negative is an addition.

Two more instrumental faults worth naming

Backlash error: wear between the screw and the nut means the spindle does not move at once when the rotation is reversed. Remedy: rotate in one direction only.

Excessive pressure: squeezing the object distorts it and the reading. Remedy: use the ratchet, which slips at a fixed torque.

Both are systematic, so more readings will not help.

06Exceptions and traps

Pitch is not least count

Pitch is the coarse advance per full turn, typically 1 mm or 0.5 mm. Least count is that figure divided by the number of circular divisions. Confusing them is the commonest error in the topic.

When rotations are mentioned, find the pitch first

'The spindle advances 2 mm in 4 rotations' means the pitch is 0.5 mm, not 2 mm. Two steps: pitch, then least count.

Different pitches can give the same least count

1 mm with 100 divisions and 0.5 mm with 50 divisions both give 0.01 mm. Never assume the standard value — compute it from the numbers the question actually gives.

The vernier division count is a number, not a length

The coinciding division number must be multiplied by the least count before it is added. Adding the bare division number produces an answer several centimetres too large.

Check the unit the options are written in

0.05 mm and 0.005 cm are the same least count. Papers routinely list one of the two and put the other value in as a distractor.

Finer is not automatically more accurate

NCERT Exercise 1.8(b) asks precisely this. Beyond a point the divisions become unreadable and random errors exceed the least count, so extra resolution buys nothing. Least count describes resolution; accuracy is a separate question.

07Numbers to remember

recognising these on sight removes a full step from most questions
Instrument / settingLeast count
Metre scale1 mm = 0.1 cm
Vernier, 10 divisions on a 1 mm main scale0.1 mm = 0.01 cm
Vernier, 20 divisions on a 1 mm main scale0.05 mm = 0.005 cm
Vernier, 25 divisions on a 1 mm main scale0.04 mm = 0.004 cm
Vernier, 50 divisions on a 1 mm main scale0.02 mm = 0.002 cm
Screw gauge, pitch 1 mm, 100 divisions0.01 mm = 10 μm
Screw gauge, pitch 1 mm, 200 divisions0.005 mm = 5 μm
Screw gauge, pitch 0.5 mm, 50 divisions0.01 mm
Screw gauge, pitch 0.5 mm, 100 divisions0.005 mm
Optical instrument resolving to one wavelength≈ 6 × 10⁻⁵ cm = 6 × 10⁻⁴ mm
Useful fractions1/20 = 0.05 · 1/25 = 0.04 · 1/50 = 0.02 · 1/200 = 0.005
Typical human hair (NCERT Ex. 1.7)≈ 0.035 mm = 35 μm

08Scientists to remember

occasionally examined in matching items; worth one reading
NameWhat to attach to the name
Pierre VernierFrench mathematician who described the sliding auxiliary scale in 1631. The instrument and the term 'vernier constant' both carry his name.
Pedro Nunes (Nonius)Portuguese mathematician whose earlier sub-division scheme gave the vernier its alternative name, the nonius.
William GascoigneFitted a micrometer screw to a telescope in the 1630s — the first use of a fine screw as a measuring device.
Jesse RamsdenBuilt the screw-cutting lathe in 1775 that made accurate micrometer screws manufacturable, and with them the modern screw gauge.
Jean Laurent PalmerPatented the handheld micrometer calliper in 1848; the screw gauge is still called the palmer in parts of Europe.

09Twenty worked questions

About the labels

Items tagged PYQ here are the NCERT chapter-end exercises, which NTA has drawn on directly. Every least count, reading and percentage on this page was recomputed by an independent script before printing.

Q01Vernier LC

A vernier callipers has 20 divisions on its sliding scale which coincide with 19 divisions of the main scale. If one main scale division is 1 mm, the least count is:

  1. 0.05 mm
  2. 0.01 mm
  3. 0.1 mm
  4. 0.005 mm
Show full solution
Given
20 VSD = 19 MSD; 1 MSD = 1 mm.
Asked
Least count of the instrument.
Concept
The least count of a vernier is the gap between one main scale division and one vernier division. That tiny difference is what lets the vernier resolve fractions of a millimetre.
Formula
LC = 1 MSD − 1 VSD = (1 MSD) / N
Baby steps
  1. 20 VSD = 19 MSD → 1 VSD = 19/20 MSD = 0.95 mm.
  2. LC = 1 MSD − 1 VSD = 1 − 0.95.
  3. = 0.05 mm = 0.005 cm.
  4. Shortcut check: LC = MSD/N = 1/20 = 0.05 mm. ✓
Answer
(a) 0.05 mm
Why not others
(a)Correct. This is the vernier described in NCERT Exercise 1.6(a).
(b)That is the least count of a standard screw gauge, not this vernier.
(c)This is the value for a 10-division vernier.
(d)Right number, wrong unit — 0.005 is the value in centimetres.
Shortcut
Whenever N vernier divisions span (N − 1) main divisions, LC = MSD / N. No subtraction needed — just divide.
Where it goes wrong
Watch the unit the options are written in. 0.05 mm and 0.005 cm are the same number; only one of them will be listed.
Q02Screw gauge LC

A screw gauge has a pitch of 1 mm and 100 divisions on its circular scale. Its least count is:

  1. 1 mm
  2. 0.1 mm
  3. 0.001 mm
  4. 0.01 mm
Show full solution
Given
Pitch = 1 mm; 100 divisions on the circular scale.
Asked
Least count.
Concept
One full rotation of the circular scale advances the spindle by one pitch. Divide that advance among the circular divisions and you get the smallest measurable step.
Formula
LC = pitch / (number of divisions on the circular scale)
Baby steps
  1. One complete rotation advances the spindle by 1 mm.
  2. That rotation is divided into 100 equal parts.
  3. So each division corresponds to 1/100 mm.
  4. LC = 0.01 mm = 10 μm = 0.001 cm.
Answer
(d) 0.01 mm
Why not others
(a)That is the pitch, not the least count.
(b)This divides by 10 instead of 100.
(c)That is the value in centimetres, not millimetres.
(d)Correct — the standard laboratory screw gauge, and the instrument in NCERT Exercise 1.6(b).
Shortcut
The standard screw gauge is 1 mm pitch, 100 divisions, LC = 0.01 mm. Carry that as your default and adjust only when the question changes the numbers.
Where it goes wrong
Do not confuse pitch with least count. The pitch is the coarse advance per turn; the least count is the fine step you can actually read.
Q03Screw gauge LC

The spindle of a screw gauge advances 2 mm in 4 complete rotations. The circular scale has 100 divisions. The least count is:

  1. 0.01 mm
  2. 0.05 mm
  3. 0.005 mm
  4. 0.002 mm
Show full solution
Given
Advance of 2 mm in 4 rotations; 100 circular divisions.
Asked
Least count.
Concept
The pitch is not given directly — you must extract it first. Pitch is the advance per single rotation.
Formula
pitch = (distance advanced) / (number of rotations) ; LC = pitch / N
Baby steps
  1. Pitch = 2 mm / 4 rotations = 0.5 mm per rotation.
  2. LC = pitch / N = 0.5 / 100.
  3. = 0.005 mm.
  4. That is half the least count of a standard screw gauge, so this instrument is twice as fine.
Answer
(c) 0.005 mm
Why not others
(a)This uses a pitch of 1 mm instead of 0.5 mm.
(b)This divides by 10 rather than 100.
(c)Correct.
(d)This comes from dividing 2 by 1000.
Shortcut
Two-step reflex whenever a question mentions rotations: find the pitch first, then divide by the number of circular divisions. Never skip straight to the division.
Where it goes wrong
Using 2 mm as the pitch. Pitch is per one rotation, and four rotations were made.
Q04Vernier reading

While measuring a rod with a vernier callipers of least count 0.01 cm, the main scale reading is 1.2 cm and the 6th vernier division coincides with a main scale division. The length of the rod is:

  1. 1.20 cm
  2. 1.26 cm
  3. 1.66 cm
  4. 1.06 cm
Show full solution
Given
MSR = 1.2 cm; coinciding vernier division = 6; LC = 0.01 cm.
Asked
Total reading.
Concept
The main scale gives the whole part; the coinciding vernier division supplies the fractional part, in units of the least count.
Formula
Reading = MSR + (VSD coinciding × LC)
Baby steps
  1. MSR = 1.2 cm.
  2. Vernier contribution = 6 × 0.01 = 0.06 cm.
  3. Total = 1.2 + 0.06.
  4. = 1.26 cm.
Answer
(b) 1.26 cm
Why not others
(a)The vernier contribution was ignored.
(b)Correct.
(c)This adds 0.46 — the vernier number was misread as a main-scale value.
(d)This subtracts the vernier reading instead of adding it.
Shortcut
The vernier reading is always added, and it is always smaller than one main scale division. If your fractional part exceeds one MSD, you have made an error.
Where it goes wrong
Multiplying by the vernier division number without the least count, giving 1.2 + 6 = 7.2 cm. The coinciding division is a count, not a length.
Q05Screw gauge reading

A screw gauge of least count 0.01 mm shows a main scale reading of 3 mm, with the 35th division of the circular scale lying on the reference line. The diameter measured is:

  1. 3.35 mm
  2. 3.035 mm
  3. 6.5 mm
  4. 3.5 mm
Show full solution
Given
Pitch-scale reading = 3 mm; circular scale division = 35; LC = 0.01 mm.
Asked
Total diameter.
Concept
Identical structure to the vernier: coarse scale plus (fine division × least count).
Formula
Reading = pitch scale reading + (circular scale division × LC)
Baby steps
  1. Main scale gives 3 mm.
  2. Circular scale gives 35 × 0.01 = 0.35 mm.
  3. Total = 3 + 0.35.
  4. = 3.35 mm.
Answer
(a) 3.35 mm
Why not others
(a)Correct.
(b)This divides 35 by 1000 instead of 100.
(c)This adds 35 × 0.1 mm.
(d)This uses 50 divisions.
Shortcut
Both instruments obey one formula: coarse reading + (fine division count × LC). Learn it once, apply it to vernier, screw gauge and spherometer alike.
Where it goes wrong
If a zero error is quoted anywhere in the question, this is not the final answer — the zero error must still be subtracted. Read to the end of the stem before answering.
Q06Zero error

A screw gauge with 100 divisions and least count 0.01 mm shows the 96th division on the reference line when its jaws are fully closed. While measuring a wire it reads 4.20 mm. The correct diameter is:

  1. 4.16 mm
  2. 4.20 mm
  3. 4.96 mm
  4. 4.24 mm
Show full solution
Given
Closed-jaw reading: 96th division; LC = 0.01 mm; observed reading = 4.20 mm.
Asked
Corrected diameter.
Concept
A closed-jaw reading near 100 means the zero mark has gone past the reference line the other way. That is a negative zero error, so the correction is positive.
Formula
zero error = −(N − n) × LC ; true = observed − zero error
Baby steps
  1. Since 96 is close to 100, the error is negative.
  2. Zero error = −(100 − 96) × 0.01 = −0.04 mm.
  3. True reading = observed − zero error = 4.20 − (−0.04).
  4. = 4.24 mm.
Answer
(d) 4.24 mm
Why not others
(a)This subtracts 0.04 instead of adding it — the sign of the zero error was missed.
(b)This ignores the zero error entirely.
(c)This treats the 96 as 96 × 0.01 = 0.96 mm added.
(d)Correct.
Shortcut
Reading near zero (say 3) when closed → positive zero error → subtract. Reading near the top (say 96) → negative zero error → add.
Where it goes wrong
Reading 96 as a positive error of 0.96 mm. The circular scale wraps around; a reading of 96 is four divisions short of a full turn, not ninety-six divisions past zero.
Q07PYQ · NCERT Ex. 1.6NCERT exercise

Which of the following is the most precise device for measuring length? (a) a vernier callipers with 20 divisions on the sliding scale, (b) a screw gauge of pitch 1 mm with 100 divisions on the circular scale, (c) an optical instrument that can measure length to within a wavelength of light.

  1. the vernier callipers
  2. the screw gauge
  3. the optical instrument
  4. all three are equally precise
Show full solution
Given
Three instruments with the specifications listed.
Asked
Which has the smallest least count.
Concept
Precision here means the smallest resolvable length. Compute all three least counts in a common unit and compare.
Formula
LCvernier = MSD/N ; LCscrew = pitch/N ; LCoptical ≈ λ
Baby steps
  1. Vernier: 1 mm / 20 = 0.05 mm = 5 × 10⁻² mm.
  2. Screw gauge: 1 mm / 100 = 0.01 mm = 1 × 10⁻² mm.
  3. Optical: a wavelength of visible light ≈ 6 × 10⁻⁵ cm = 6 × 10⁻⁴ mm.
  4. The optical instrument's least count is over ten times smaller than the screw gauge's, so it is the most precise.
Answer
(c) the optical instrument
Why not others
(a)The coarsest of the three, at 0.05 mm.
(b)Ten times finer than the vernier, but still far coarser than a wavelength of light.
(c)Correct — roughly 6 × 10⁻⁴ mm.
(d)Their least counts differ by more than two orders of magnitude.
Shortcut
Order to carry into the exam, coarsest to finest: metre scale 1 mm, vernier 0.1–0.05 mm, screw gauge 0.01 mm, optical/interference methods 10⁻⁴ mm.
Where it goes wrong
Converting all three to the same unit is the whole difficulty. Wavelengths quoted in Ångströms or nanometres must be brought to millimetres before comparing.
Q08PYQ · NCERT Ex. 1.7NCERT exercise

A student measures the thickness of a human hair through a microscope of magnification 100. Twenty observations give an average width of 3.5 mm in the field of view. The estimated thickness of the hair is:

  1. 350 mm
  2. 0.035 mm
  3. 3.5 mm
  4. 0.35 mm
Show full solution
Given
Magnification = 100; observed average width = 3.5 mm.
Asked
Actual thickness of the hair.
Concept
Magnification is the ratio of image size to object size, so the real object is the observed size divided by the magnification.
Formula
magnification = image size / object size
Baby steps
  1. Observed (image) width = 3.5 mm.
  2. Object size = image size / magnification = 3.5 / 100.
  3. = 0.035 mm = 3.5 × 10⁻⁵ m = 35 μm.
  4. That is the right order for a human hair, which runs roughly 20–100 μm.
Answer
(b) 0.035 mm
Why not others
(a)This multiplies by 100 instead of dividing.
(b)Correct.
(c)This ignores the magnification altogether.
(d)This divides by 10 instead of 100.
Shortcut
A magnifying instrument always makes things look bigger, so the true value must come out smaller. If your answer is larger than the observed value, you have divided the wrong way.
Where it goes wrong
The twenty observations are there only to justify averaging — they do not enter the arithmetic. Extra data in a stem is often there to test whether you know what is relevant.
Q09Vernier LC

A vernier scale has 10 divisions coinciding with 9 divisions of the main scale, where one main scale division is 1 mm. The vernier constant in centimetres is:

  1. 0.01 cm
  2. 0.1 cm
  3. 0.001 cm
  4. 0.05 cm
Show full solution
Given
10 VSD = 9 MSD; 1 MSD = 1 mm.
Asked
Vernier constant (least count) in centimetres.
Concept
'Vernier constant' is simply another name for the least count of a vernier callipers.
Formula
LC = MSD / N
Baby steps
  1. LC = 1 mm / 10 = 0.1 mm.
  2. Convert: 0.1 mm = 0.01 cm.
  3. So the vernier constant is 0.01 cm.
  4. This is the commonest school-laboratory vernier.
Answer
(a) 0.01 cm
Why not others
(a)Correct.
(b)That is the value in millimetres, not centimetres.
(c)One order of magnitude too small.
(d)That belongs to a 20-division vernier.
Shortcut
Two verniers cover almost every question: 10 divisions → 0.01 cm, and 20 divisions → 0.005 cm. Recognise which one you have been given and the LC is immediate.
Where it goes wrong
The words 'vernier constant', 'least count' and 'vernier scale division difference' all mean the same thing. Different papers use different names for it.
Q10PYQ · NCERT Ex. 1.8(b)NCERT exercise

A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Is it possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

  1. Yes, accuracy improves without limit
  2. Yes, but only up to 1000 divisions
  3. Yes, provided the pitch is increased in the same proportion
  4. No — beyond a point the divisions become too crowded to read, and random errors dominate the finer least count
Show full solution
Given
Pitch 1.0 mm; 200 circular divisions, giving LC = 0.005 mm.
Asked
Whether accuracy can be raised without limit this way.
Concept
Least count is a statement about the instrument's resolution. Accuracy also depends on whether a human can read that resolution and on the random errors already present in the measurement.
Formula
LC = pitch / N → 0 as N → ∞, but accuracy does not
Baby steps
  1. Arithmetically, LC = 1.0/200 = 0.005 mm, and larger N gives a smaller LC.
  2. But the circular scale has a fixed circumference, so more divisions means finer lines.
  3. Past a point they cannot be told apart by eye, and the reading becomes a guess.
  4. Random errors — temperature, screw wear, uneven pressure on the ratchet — also exceed the least count, so the extra resolution is meaningless. Accuracy cannot be improved arbitrarily.
Answer
(d) No — beyond a point the divisions become too crowded to read, and random errors dominate the finer least count
Why not others
(a)Resolution improves on paper, but readability and random errors set a ceiling.
(b)There is no such sharp numerical cut-off; the limit is practical, not a fixed number.
(c)Increasing the pitch in proportion would leave the least count unchanged, defeating the purpose.
(d)Correct.
Shortcut
Least count sets the floor on the uncertainty of a single reading. Once random errors exceed that floor, a finer instrument buys nothing.
Where it goes wrong
This is a reasoning question, not a calculation. NEET has asked several statement items built on exactly this distinction between resolution and accuracy.
Q11Zero error

When the jaws of a vernier callipers of least count 0.01 cm are closed, the 3rd vernier division coincides with a main scale division. A rod then reads 2.34 cm. Its true length is:

  1. 2.37 cm
  2. 2.34 cm
  3. 2.31 cm
  4. 2.30 cm
Show full solution
Given
Closed-jaw coincidence at the 3rd division; LC = 0.01 cm; observed reading = 2.34 cm.
Asked
True length of the rod.
Concept
A closed-jaw reading of 3 divisions means the instrument reports 0.03 cm too much on every measurement. That is a positive zero error and must be subtracted.
Formula
zero error = n × LC ; true = observed − zero error
Baby steps
  1. Zero error = 3 × 0.01 = +0.03 cm.
  2. True length = observed − zero error.
  3. = 2.34 − 0.03.
  4. = 2.31 cm.
Answer
(c) 2.31 cm
Why not others
(a)This adds the zero error instead of subtracting it.
(b)This ignores the zero error.
(c)Correct.
(d)This subtracts 0.04 cm.
Shortcut
Small closed-jaw reading counted upward from zero → positive error → subtract. This is the vernier twin of the screw-gauge question above.
Where it goes wrong
For a vernier, a negative zero error shows up as a coincidence counted backwards from the last division, exactly as with the screw gauge. Check which side of zero the vernier's zero mark sits on.
Q12Screw gauge LC

A screw gauge has a pitch of 1.0 mm and 200 divisions on its circular scale. Its least count is:

  1. 0.01 mm
  2. 0.005 mm
  3. 0.002 mm
  4. 0.05 mm
Show full solution
Given
Pitch = 1.0 mm; 200 circular divisions.
Asked
Least count.
Concept
Same formula as before, with a finer circular scale.
Formula
LC = pitch / N
Baby steps
  1. LC = 1.0 mm / 200.
  2. = 0.005 mm.
  3. = 5 μm, half the standard screw gauge's least count.
  4. This is the instrument described in NCERT Exercise 1.8(b).
Answer
(b) 0.005 mm
Why not others
(a)That is the 100-division instrument.
(b)Correct.
(c)This divides by 500.
(d)This divides by 20.
Shortcut
Doubling the number of divisions halves the least count. The relationship is strictly inverse, which is why the NCERT question about increasing divisions is interesting at all.
Where it goes wrong
A smaller least count means better resolution but not automatically better accuracy — see the previous question.
Q13Vernier LC

N divisions of a vernier scale coincide with (N − 1) divisions of the main scale. If one main scale division is s, the least count of the vernier is:

  1. s / N
  2. s / (N − 1)
  3. (N − 1)s / N
  4. N s
Show full solution
Given
N VSD = (N − 1) MSD; 1 MSD = s.
Asked
Least count in terms of s and N.
Concept
Derive the general result rather than memorising it — then any set of numbers becomes a one-line substitution.
Formula
LC = 1 MSD − 1 VSD
Baby steps
  1. N VSD = (N − 1) MSD → 1 VSD = [(N − 1)/N] s.
  2. LC = 1 MSD − 1 VSD = s − [(N − 1)/N] s.
  3. = s[1 − (N − 1)/N] = s[(N − N + 1)/N].
  4. = s / N.
Answer
(a) s / N
Why not others
(a)Correct — this single formula generates every vernier least count in the chapter.
(b)Comes from dividing by (N − 1) rather than N.
(c)This is the size of one vernier division, not the least count.
(d)This multiplies where it should divide, and would make the vernier coarser than the main scale.
Shortcut
Check the derived formula against a known case: N = 10, s = 1 mm gives 0.1 mm, which matches the standard vernier. A formula that survives a known case is almost certainly right.
Where it goes wrong
Some papers state the relation as N VSD = (N + 1) MSD instead. Then LC = s/N still, but one vernier division is larger than a main division — a backward vernier. Re-derive rather than recall if the wording changes.
Q14Precision

A screw gauge of least count 0.01 mm is used to measure the diameter of a wire, and the reading is 3.24 mm. The percentage error in the measurement is approximately:

  1. 0.1%
  2. 1%
  3. 3.1%
  4. 0.31%
Show full solution
Given
LC = 0.01 mm; measured diameter = 3.24 mm.
Asked
Percentage error.
Concept
For a single reading, the uncertainty is taken as one least count. The percentage error is that uncertainty as a fraction of the reading.
Formula
percentage error = (LC / measured value) × 100
Baby steps
  1. Absolute uncertainty = 0.01 mm.
  2. Relative error = 0.01 / 3.24 = 0.003086.
  3. × 100 = 0.3086%.
  4. ≈ 0.31%.
Answer
(d) 0.31%
Why not others
(a)This would need a reading of about 10 mm.
(b)This would need a reading of about 1 mm.
(c)Two orders of magnitude too large.
(d)Correct.
Shortcut
The same least count gives a smaller percentage error on a thicker object. This is the argument for measuring a bundle of wires and dividing, rather than measuring one wire.
Where it goes wrong
Do not halve the least count. Some textbooks quote the uncertainty of a single reading as ±½ LC; NEET consistently uses one full least count unless told otherwise.
Q15Instrument errors

Backlash error in a screw gauge arises because:

  1. the zero of the circular scale does not coincide with the reference line
  2. the observer's eye is not in line with the scale
  3. on reversing the direction of rotation, the screw does not begin to move immediately
  4. the wire is compressed too tightly between the jaws
Show full solution
Given
Definition of backlash error.
Asked
Its cause.
Concept
Wear between the screw and the nut leaves a small gap. When the rotation is reversed, that gap must be taken up before the spindle actually moves, so the scale turns while the spindle stands still.
Formula
avoid by rotating in one direction only
Baby steps
  1. Wear creates play between the screw threads and the nut.
  2. Reversing the rotation first closes that play.
  3. During those few divisions the circular scale moves but the spindle does not.
  4. Remedy: always turn the screw in one direction only while taking a set of readings.
Answer
(c) on reversing the direction of rotation, the screw does not begin to move immediately
Why not others
(a)That is zero error, a different instrumental fault.
(b)That is parallax error, an observational fault.
(c)Correct.
(d)That is an error from excessive pressure, which is why a ratchet is fitted.
Shortcut
Three named screw-gauge faults, three different remedies: zero error → subtract it; backlash → rotate one way only; excessive pressure → use the ratchet.
Where it goes wrong
All three are systematic errors, so none of them is reduced by taking more readings. Only correction or better technique helps.
Q16Precision

A vernier callipers of least count 0.01 cm measures a rod of length 4.35 cm. The relative error in the measurement is closest to:

  1. 0.023%
  2. 0.23%
  3. 2.3%
  4. 0.0023%
Show full solution
Given
LC = 0.01 cm; length = 4.35 cm.
Asked
Relative error as a percentage.
Concept
Divide the least count by the measured value, then multiply by 100.
Formula
percentage error = (LC / value) × 100
Baby steps
  1. 0.01 / 4.35 = 0.0022989.
  2. × 100 = 0.22989%.
  3. ≈ 0.23%.
  4. Note the corresponding fraction, 0.0023, is the relative error before conversion.
Answer
(b) 0.23%
Why not others
(a)One order of magnitude too small.
(b)Correct.
(c)One order of magnitude too large.
(d)This is the relative error expressed as a fraction, not a percentage.
Shortcut
Options in this style are usually spaced by factors of ten, so the whole question reduces to placing the decimal point correctly. Do the division first, then decide the power of ten.
Where it goes wrong
Options (a), (b), (c) and (d) here are the same digits with four different exponents. Reading 'relative error' as a fraction when the question wants a percentage is exactly what option (d) is fishing for.
Q17Vernier LC

The main scale of a vernier callipers is graduated in millimetres, and 25 vernier divisions coincide with 24 main scale divisions. The least count is:

  1. 0.04 mm
  2. 0.025 mm
  3. 0.4 mm
  4. 0.004 mm
Show full solution
Given
25 VSD = 24 MSD; 1 MSD = 1 mm.
Asked
Least count.
Concept
This is the general N and (N − 1) pattern with N = 25.
Formula
LC = MSD / N
Baby steps
  1. 1 VSD = 24/25 mm = 0.96 mm.
  2. LC = 1 − 0.96 = 0.04 mm.
  3. Check with the shortcut: 1/25 = 0.04 mm. ✓
  4. In centimetres this is 0.004 cm.
Answer
(a) 0.04 mm
Why not others
(a)Correct.
(b)This comes from 1/40, not 1/25.
(c)One order of magnitude too large.
(d)That is the value in centimetres.
Shortcut
1/25 = 0.04 is worth recognising instantly, as is 1/20 = 0.05 and 1/50 = 0.02. These three cover almost every vernier NEET has printed.
Where it goes wrong
Computing 24/25 = 0.96 and then answering 0.96 mm. That is the size of one vernier division, not the least count — you still have to subtract it from 1 mm.
Q18Instrument choice

The most suitable instrument for measuring the diameter of a thin wire of about 0.5 mm is:

  1. a metre scale
  2. a vernier callipers
  3. a spherometer
  4. a screw gauge
Show full solution
Given
Object diameter ≈ 0.5 mm.
Asked
The appropriate instrument.
Concept
Choose the instrument whose least count is small compared with the quantity being measured — roughly a hundredth of it or better.
Formula
compare LC with the size of the object
Baby steps
  1. Metre scale: LC 1 mm — larger than the object itself. Useless.
  2. Vernier: LC 0.1 mm — only five divisions across the wire, giving 20% resolution.
  3. Screw gauge: LC 0.01 mm — fifty divisions across the wire, about 2%. Suitable.
  4. So the screw gauge is the right choice.
Answer
(d) a screw gauge
Why not others
(a)Its least count exceeds the object's diameter.
(b)Usable but coarse — only about 20% resolution on a 0.5 mm wire.
(c)A spherometer measures the thickness of plates and the radius of curvature of surfaces, not wire diameters.
(d)Correct — this is the standard laboratory use of the screw gauge.
Shortcut
Rough guide by scale: metre scale for centimetres, vernier for millimetres, screw gauge for tenths of a millimetre, spherometer for the curvature of surfaces.
Where it goes wrong
A spherometer also has a pitch and a circular scale and looks similar in a formula sheet, but it is for surfaces. Match the instrument to the geometry, not just to the least count.
Q19PYQ · NCERT Ex. 1.8(a)NCERT exercise

You are given a thread and a metre scale. The best way to estimate the diameter of the thread is:

  1. measure the thread directly against the metre scale
  2. fold the thread in half repeatedly and measure
  3. wind n closely spaced turns of the thread around a pencil, measure the length L of the coil, and take the diameter as L/n
  4. weigh the thread and use its density
Show full solution
Given
A thread and a metre scale of least count 1 mm.
Asked
A method to estimate the thread's diameter.
Concept
A single thread is far thinner than the metre scale's least count. Stacking many identical thicknesses side by side scales the quantity up into the measurable range, and dividing at the end scales the answer back down — while dividing the error by n as well.
Formula
d = L / n
Baby steps
  1. Wind n turns of the thread tightly and without gaps around a cylindrical rod.
  2. Measure the total length L of the wound section with the metre scale.
  3. Each turn contributes one diameter, so d = L/n.
  4. The 1 mm least count is shared among n turns, so the error in d is (1 mm)/n — the same idea as timing 20 oscillations of a pendulum.
Answer
(c) wind n closely spaced turns of the thread around a pencil, measure the length L of the coil, and take the diameter as L/n
Why not others
(a)The diameter is far below the 1 mm least count; a direct reading is meaningless.
(b)Folding changes the length, not the diameter.
(c)Correct — the multiplication method described in NCERT.
(d)Density and cross-section are not given, and this introduces more unknowns than it removes.
Shortcut
Whenever a quantity is smaller than the least count, look for a way to measure n of them at once. It is the same trick behind timing many oscillations and weighing many identical objects together.
Where it goes wrong
The turns must be closely wound with no gaps. Any spacing between turns inflates L and therefore the estimated diameter — a systematic error, not a random one.
Q20Screw gauge reading

A screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale. While measuring a wire, the main scale reads 2.5 mm and the 20th division of the circular scale is on the reference line. The diameter of the wire is:

  1. 2.50 mm
  2. 2.70 mm
  3. 4.50 mm
  4. 2.52 mm
Show full solution
Given
Pitch = 0.5 mm; 50 circular divisions; MSR = 2.5 mm; CSR = 20.
Asked
Diameter of the wire.
Concept
Compute the least count first, then apply the standard reading formula. Two steps, always in that order.
Formula
LC = pitch/N ; reading = MSR + (CSR × LC)
Baby steps
  1. LC = 0.5 / 50 = 0.01 mm.
  2. Circular contribution = 20 × 0.01 = 0.20 mm.
  3. Reading = 2.5 + 0.20.
  4. = 2.70 mm.
Answer
(b) 2.70 mm
Why not others
(a)The circular scale contribution was omitted.
(b)Correct.
(c)This adds 20 × 0.1 mm.
(d)This uses an LC of 0.001 mm.
Shortcut
Note that a 0.5 mm pitch with 50 divisions gives the same 0.01 mm least count as a 1 mm pitch with 100 divisions. Different instruments, identical resolution.
Where it goes wrong
Skipping the least-count step and assuming 0.01 mm out of habit happens to work here, but it will not when the numbers change. Always compute the LC from the pitch actually given.